冒泡排序 Python

Question

我正在尝试在 Python 中进行冒泡排序，但不创建函数、导入函数等

到目前为止我已经明白了，但现在我被难住了:p

array = [1, 42, 321, 44, 121, 61, 812, 71, 10, 11]
number = 1
ArrayIndex = 0
numchange = 0
TotalNumberofLoops = 10
OuterLoop = 0
InnerLoop = 0

while OuterLoop < TotalNumberofLoops:
  InnerLoop = OuterLoop + 1
  while InnerLoop < TotalNumberofLoops:
    if array[OuterLoop] < array[InnerLoop]:
      numchange = array[InnerLoop]
      array[OuterLoop] = array[InnerLoop]
      array[InnerLoop] = numchange

    InnerLoop=InnerLoop + 1
  print array
  OuterLoop = OuterLoop + 1

这给出了以下输出：

[812, 42, 321, 44, 121, 61, 812, 71, 10, 11]
[812, 812, 321, 44, 121, 61, 812, 71, 10, 11]
[812, 812, 812, 44, 121, 61, 812, 71, 10, 11]
[812, 812, 812, 812, 121, 61, 812, 71, 10, 11]
[812, 812, 812, 812, 812, 61, 812, 71, 10, 11]
[812, 812, 812, 812, 812, 812, 812, 71, 10, 11]
[812, 812, 812, 812, 812, 812, 812, 71, 10, 11]
[812, 812, 812, 812, 812, 812, 812, 71, 10, 11]
[812, 812, 812, 812, 812, 812, 812, 71, 11, 11]
[812, 812, 812, 812, 812, 812, 812, 71, 11, 11]

感谢任何解决方案！

Answer 1

试试这个：

array = [1, 42, 321, 44, 121, 61, 812, 71, 10, 11]
number = 1
ArrayIndex = 0
numchange = 0
TotalNumberofLoops = 10
OuterLoop = 0
InnerLoop = 0

while OuterLoop < TotalNumberofLoops:
    InnerLoop = OuterLoop + 1
    while InnerLoop < TotalNumberofLoops:
        if array[OuterLoop] < array[InnerLoop]:
            array[OuterLoop], array[InnerLoop] = array[InnerLoop], array[OuterLoop]

        InnerLoop = InnerLoop + 1
    print(array)
    OuterLoop = OuterLoop + 1

这种方式的元素交换更加 pythonic 并且也是正确的。

Answer 2

怎么样

   def bubble_sort(x):
        # bubble sort with early termination - O(N) for nearly sorted
        swapped = False
        if len(x) > 1:
            for i in range(len(x) - 1):
                    if x[i] > x[i + 1]:
                        swapped = True
                        x[i+1], x[i] = x[i], x[i+1]
            if swapped:
                return bubble_sort(x[0:-1])+[x[-1]]
        return x

冒泡排序 Python

Bubble Sort Python

python

bubble-sort